Question

Решите задания(желательно с областью определения)

Answer 1

$1)quad y= frac{x}{x^3+x} \\OOF:; ; x^3+xne 0; ,\\x(x^2+1)ne 0; ; Rightarrow ; ; xne 0\\xin (-infty ,0)cup (0,+infty )\\2)quad y=arcsinfrac{1+x^2}{2x}$

$OOF:; ; -1 leq frac{1+x^2}{2x} leq 1\\ left { {{ frac{1+x^2}{2x} leq 1 } atop {frac{1+x^2}{2x} geq -1}} right. ; left { {{frac{1+x^2-2x}{2x} leq 0} atop {frac{1+x^2+2x}{2x} geq 0}} right. left { {{frac{(x-1)^2}{2x} leq 0} atop {frac{(x+1)^2}{2x} geq 0}} right. ; left { {{x textless 0,x=1} atop {x textgreater 0,x=-1}} right. ; to x={-1;1}$

$3)quad y=sqrt{2^{x}-3^{x}}\\OOF:; ; 2^{x}-3^{x} geq 0\\2^{x} geq 3^{x}; |:3^{x} textgreater 0\\(frac{2}{3})^{x} geq 1\\(frac{2}{3})^{x} geq (frac{2}{3})^0\\x leq 0\\xin (-infty ,0, ]$

$4)quad y=0,5^{sqrt{4-x^2}}+frac{1}{x+1}\\OOF:; ; left{begin{array}{cc}4-x^2& geq 0\x+1&ne 0end{array}right \\a); ; 4-x^2 geq 0; ,; ; (2-x)(2+x) geq 0; ,; ; (x-2)(x+2) leq 0\\xin [-2,2, ]\\b); ; x+1ne 0; ,; ; xne -1\\ left { {{xin [-2,2, ]} atop {xne -1}} right. \\xin [-2,-1)cup (-1,2, ]\\5)quad y=frac{1}{sqrt{14+5x-x^2}}+sqrt{x^2-x-20}$

$a); ; x^2-x-20 geq 0; ,; ; (x-5)(x+4) geq 0\\xin (-infty ,-4, ]cup [, 5,+infty )$

$b); ; 14+5x-x^2 textgreater 0\\x^2-5x-14 textless 0\\(x-7)(x+2) textless 0quad +++(-2)---(7)+++\\xin (-2,7)\\c); ; left { {{xin (-infty ,-4, ]cup [, 5,+infty )} atop {xin (-2,7)}} right. ; ; to ; ; xin [, 5,7)$